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In my class we're learning vector spaces, and in the text book there's an example with no solution and it goes like this:

If the domain of functions $f$ and $g$ is $[-1,1]$ and if they are defined $f(x) = \arcsin\left(\displaystyle\frac{2x}{1+x^2}\right)$ and $g(x) = \arctan(x)$, then $(f,g)$ is linearly independent?

I don't know how to prove this, if I can make a linear combination of one of them using the other it's dependent, but how should I go about doing that?

Thanks in advance.

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3 Answers

If you want to show that $f$ and $g$ are linearly independent, a very similar (but different) question was asked recently. Many techniques for solving that problem are also applicable here.

If you want to show that $f$ and $g$ are linearly dependent, you have to show that $\arcsin(2x/(1+x^2))$ is some constant multiple of $\arctan x$.

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Theorem: Let $f$ be a function defined in an intrval $I$, and let $F$ be an antiderivative of $f$ in $I$. Then every other derivative of $f$ in $I$ is of the form $F+costant$.

Now use above theorem to see that: $$\arcsin\left(\frac{2x}{1+x^2}\right)= \left\{ \begin{array}{ll} -\pi-2\arctan(x) & \quad x \leq-1 \\ 2\arctan(x) & \quad -1\leq x\leq1 \\\pi-2\arctan(x) & \quad x\geq1 \end{array} \right. $$

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So between [-1,1] it's linearly dependent, and [1, infinity] linearly independent. –  user55348 Jan 6 '13 at 13:06
    
@user55348: Yes. Try to look for the conditions which two functions are linierely dependent. See user1551's answer for the link or google it. –  B. S. Jan 6 '13 at 14:59
    
good lesson taught! +1 –  amWhy Feb 22 '13 at 2:07
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x=tan g(x) sin f(x)=2tan g(x)/1+tan^2g(x)=sin 2g(x) so f(x)=2g(x) (under suitable conditions.

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